Question

If
$\alpha$
and
$\beta$
are the zeroes of the polynomial P (x) = x^2 + 2x + 3, then value of 1/
$\alpha$
+ 1/
$\beta$
is
A] -2/3
B] -3/2
C] 3/5
D] 2/3​

Answer 1

Answer :

The correct option is :

(A) -2/3

Given :

The quadratic polynomial is :

• x² + 2x + 3
• α and ß are the zeroes of the polynomial

To Find :

• $\sf{\dfrac{1}{\alpha} + \dfrac{1}{\beta}}$

Concept to be used :

The relationship between the zeroes and the coefficients of the polynomial :

$\sf \star \: {sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }$

$\star \: \sf{product \: of \: the \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } }$

Solution :

Using the relationship of zeroes and the coefficients in the given polynomial :

Sum of the zeroes :

$\sf \alpha + \beta = - \dfrac{ 2}{1} \\ \\ \sf \implies \alpha + \beta = - 2 \: \: ...........(1)$

Now Product of the zeroes :

$\sf \alpha \beta = \dfrac{3}{1} \\ \\ \sf \implies \alpha \beta = 3 \: \: ............(2)$

Dividing equation (1) by (2) we have :

$\sf \implies \dfrac{ \alpha + \beta }{ \alpha \beta } = \dfrac{ - 2}{3} \\ \\ \implies \sf \frac{ \alpha }{ \alpha \beta } + \frac{ \beta }{ \alpha \beta } = \frac{ - 2}{3} \\ \\ \sf \implies \dfrac{1}{ \beta } + \dfrac{1}{ \alpha } = \frac{ - 2}{3} \\ \\ \bf \implies \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \frac{ - 2}{3}$

Answer 2

Answer:

The correct option is :

(A) -2/3

Given :

The quadratic polynomial is :

x² + 2x + 3

alpha and ß are the zeroes of the polynomial

To Find :

Concept to be used :

The relationship between the zeroes and the coefficients of the polynomial :

Solution :

Using the relationship of zeroes and the coefficients in the given polynomial :

Sum of the zeroes :

Now Product of the zeroes :

Dividing equation (1) by (2) we have :

Step-by-step explanation:

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